Richard Bass' Real Analysis Exercise 3.2 I am self-studying from Richard Bass's Real Analysis book, available here. 
I am not sure about Exercise 3.2:

Suppose $(X,\mathcal{A})$ is a measurable space and $\mu$ is a
  non-negative set function that is finitely additive and such that
  $\mu(\varnothing) = 0$ and $\mu(X)<\infty$. Suppose that whenever $A_i$ is a sequence of sets in $\mathcal{A}$ which decrease to $\varnothing$, then $\lim_{i\to\infty}\mu(A_i)=0$. Show that $\mu$ is a measure.

In order to show that $\mu$ is a measure, all that remains to be shown is for any countable collection of pairwise disjoint sets $\{B_i\}_{ i =1,2,3\ldots}$, 
$$\mu(\cup_i B_i)=\sum_i \mu(B_i).$$
My attempt:
Let $\{B_i\}_{ i =1,2,3\ldots}$ be a countable collection of pairwise disjoint sets. Then consider $\cup_{i=n}^{\infty}B_i \downarrow \varnothing$. It follows that $\lim_{n\to\infty}\mu(\cup_{i=n}^{\infty}B_i)=0$. We can rewrite this as
$$\lim_{n\to\infty}\mu\left(\cup_{i=1}^{\infty}B_i-\cup_{i=1}^{n-1}B_i\right)=0,$$
so by the algebra of limits
$$\mu\left(\cup_{i=1}^{\infty}B_i\right)=\lim_{n\to\infty}\mu(\cup_{i=1}^{n-1}B_i)=\lim_{n\to\infty}\mu(\cup_{i=1}^n B_i)=\lim_{n\to\infty}\sum_{i=1}^n\mu(B_i)=\sum_{i=1}^{\infty}\mu(B_i).$$
Questions:
(1) Is this proof correct? (In particular I am not sure that $\cup_{i=n}^{\infty}B_i$ does decrease to $\varnothing$.)
(2) I do not appear to have used $\mu(X)<\infty$ anywhere. Why is this needed?
 A: You cannot write $\mu (A\setminus B)$ as $\mu (A)-\mu (B)$ for infinite measures  (for $B \subset A$).  That is where the assumption $\mu (X) <\infty $ is required. Otherwise your proof is correct. It should be clear that $\cup_{i=n}^{\infty }B_i$ is decreasing. Suppose there is a point $x$ in the intersection. Then $x$ would belong to infinitely many of the sets $B_i$. But no point can belong to more than one of these sets, so the intersection is empty. 
A: As Kavi Rama Murthy already noted, you use $\mu(X) < \infty$ when you write $\mu(A\setminus B) = \mu(A)-\mu(B)$. Actually, this is a stronger than what is needed: For $B\subseteq A$, it suffices that $\mu(B) < \infty$. This is easy to see from the proof: we have $A = (A\setminus B) \cup B$, and the two sets on the right hand side are disjoint, so we have $$\mu(A) = \mu(A\setminus B) + \mu(B).$$ 
Now in order to be able to subtract $\mu(B)$ from both sides of this equation, you need $\mu(B) < \infty$. Of course, the most natural assumption that guarantees this for arbitrary measurable sets is $\mu(X) < \infty$.
